Is object a formula?
is_formula()
tests if x
is a call to ~
. is_bare_formula()
tests in addition that x
does not inherit from anything else than
"formula"
.
is_formula(x, scoped = NULL, lhs = NULL) is_bare_formula(x, scoped = NULL, lhs = NULL)
x |
An object to test. |
scoped |
A boolean indicating whether the quosure is scoped,
that is, has a valid environment attribute. If |
lhs |
A boolean indicating whether the formula
or definition has a left-hand side. If |
The scoped
argument patterns-match on whether the scoped bundled
with the quosure is valid or not. Invalid scopes may happen in
nested quotations like ~~expr
, where the outer quosure is validly
scoped but not the inner one. This is because ~
saves the
environment when it is evaluated, and quoted formulas are by
definition not evaluated.
x <- disp ~ am is_formula(x) is_formula(~10) is_formula(10) is_formula(quo(foo)) is_bare_formula(quo(foo)) # Note that unevaluated formulas are treated as bare formulas even # though they don't inherit from "formula": f <- quote(~foo) is_bare_formula(f) # However you can specify `scoped` if you need the predicate to # return FALSE for these unevaluated formulas: is_bare_formula(f, scoped = TRUE) is_bare_formula(eval(f), scoped = TRUE)
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